Proving no packing can beat a number — with a magic function and its Fourier transform
Suppose you want to know how densely spheres can fill space. There are two halves to that question and they are not remotely the same difficulty. To show that at least 74% of space can be filled, you build a stack of oranges and point at it — one arrangement, exhibited, done. To show that no more than 74% can be filled, you have to rule out every arrangement there is: every lattice, every periodic packing, every lopsided, aperiodic, deliberately perverse configuration of infinitely many spheres that anyone will ever think of, and all the ones nobody will.
A lower bound is a construction. An upper bound is an argument about objects you have never seen. And the single most powerful such argument anyone has found is startlingly compact: exhibit one radial function with two sign conditions, and the whole infinite space of packings in that dimension collapses underneath it.
Before any Fourier analysis, look at the scoreboard. For each dimension there is a floor — the densest packing anyone has built — and a ceiling — the smallest number anyone has proved impossible to beat. The true density is one specific number trapped between them, and in all but five dimensions nobody knows which. Dimension 13 is typical: somebody built a packing filling 3.20% of space, somebody else proved nothing fills more than 6.26%, and the honest answer to “how dense is dimension 13?” is a shrug spanning a factor of two.
The band is closed. Viazovska 2016 produced a proof that nothing beats 25.37%, and the packing that achieves it was already on the table.
Hover the strip or press Sweep. Five columns sit flat on the baseline — dimensions 1, 2, 3, 8, 24 — and every other one stands up. Dimension 19 is the worst offender at 2.40×. Upper bounds for the unsolved dimensions are the Cohn–Elkies linear programming values published in 2003, still essentially the state of the art; later refinements move them by a percent or so.
Try it: Press Sweep 1 → 24 and watch the band. It is shut in dimensions 1, 2 and 3, blows open through the middle dimensions, slams shut at 8, opens again immediately, and closes once more at 24. The strip at the bottom is the same story on one axis: five columns flat on the floor, nineteen standing up. Dimension 3 hides a detail worth noticing — the bound that closed it was Hales' exhaustive case analysis, not the method in this lesson, which stalls at ≈ 77.98% against a true 74.05%.
In 2003, Henry Cohn and Noam Elkies published a theorem in the Annals of Mathematics that reads like a conjuring trick. Take any radial function f on n-dimensional space, not identically zero, and ask three things of it:
Then no packing of spheres of radius r/2 anywhere in that dimension has density greater than vol(Br/2) · f(0) / f̂(0). One function, three lines of hypothesis, and an infinite family of arrangements is bounded — including all the irregular ones. The reason it works is Poisson summation, which turns a sum over a packing into a sum over frequencies, and the two sign conditions then attack the two sides of that identity from opposite directions.
Two totals, computed independently over two different sets of points, and they agree to every digit shown: 1.285541 on the left, 1.285541 on the right. That is Poisson summation, and it is the only fact about lattices this argument needs.
The dot at each lattice point is sized by how much that point contributes. Violet is positive, rose is negative, cyan is the Fourier side. The lattice shown is the hexagonal one — the densest packing in the plane.
This is the argument for lattice packings, where Poisson summation applies directly. Real packings need not be lattices, and the general proof is genuinely harder: Cohn and Elkies handle it by taking periodic packings with many spheres per cell — which come arbitrarily close to any packing — and running a version of this calculation with an averaged sum over the translates. The conclusion is identical, and so is the pair of sign conditions.
Try it: Step through the argument. The two totals in step 1 are computed separately — one over the lattice, one over its dual — and they agree to six decimal places, which is Poisson summation being true in front of you. Then watch condition ① knock every off-origin term negative, condition ② forbid a single negative term on the other side, and the two inequalities close on the covolume from both ends. Switch between the hexagonal, square and stretched lattices: the left-hand density changes every time and the right-hand ceiling never does.
The theorem says a good function certifies a good bound. It does not say how to find one, and that is the entire difficulty. Below is the method with the lid off. The sliders build f out of Laguerre functions — chosen because each one is an eigenfunction of the n-dimensional Fourier transform with eigenvalue (−1)k, so f̂ comes out exactly rather than by numerical integration. Move a slider and both curves respond at once; the two sign conditions light up green or red; and when both hold, the number underneath is a theorem.
One structural fact falls out before you touch anything. For large radii the highest-order term dominates, and it fixes the sign of f's tail as well as f̂'s. If the leading nonzero coefficient sits at an even index, the two conditions demand that the same number be ≤ 0 and ≥ 0 simultaneously — no tuning survives that. The leading term has to be an odd one, always.
Both conditions green means a theorem: no packing in 8 dimensions, lattice or otherwise, exceeds that percentage. The sliders move a combination of Laguerre functions, chosen because each is its own Fourier transform up to sign — so f̂ is computed exactly rather than numerically. Getting from here to the exact answer in dimensions 8 and 24 took functions built from modular forms, with double roots at every distance occurring in the lattice, and thirteen years of searching.
Try it: Start in dimension 8 with the two-term optimum and push c₃ up until f̂ dips below the axis — the certificate dies instantly, and the bound with it. Then zero out c₁ and watch the even-leading-term argument bite. The honest scorecard for the default function: it is about 10% above the truth in dimension 2, about 2× off in dimension 8, and off by a factor of ≈ 19.5 by dimension 24. Every one of those numbers is a valid theorem and a loose one.
That looseness is the point of the rest of this lesson. Cohn and Elkies could compute far better functions than four sliders allow — with a few dozen terms and real optimisation their bounds were the best known in every dimension from 4 to 36, and they still essentially are. What they could not do was reach equality. Numerically, in dimensions 8 and 24, the optimum looked exactly like E₈ and the Leech lattice to thirty decimal places. The functions that would prove it stayed out of reach for thirteen years.
Ask what it would take for the bound to be not just true but exactly right. A chain of inequalities is tight only when every link is an equality, and tracing that back through Poisson summation produces a merciless demand: f must vanish at every distance that actually occurs between points of the lattice you are trying to certify. For E₈ that list is √2, √4, √6, √8, … forever. And because f is already required to be ≤ 0 out there, vanishing at an interior point means touching zero from below — a double root at every one of those radii, with f̂ doing the same thing from above.
That is not a function you write down by guessing. In March 2016 Maryna Viazovska constructed one explicitly for dimension 8, assembling it from modular forms whose quasi-periodicity in the upper half-plane becomes, after an integral transform, exactly the required pattern of zeros. Within a week she and Henry Cohn, Abhinav Kumar, Stephen D. Miller and Danylo Radchenko had done dimension 24. E₈ fills π⁴/384 ≈ 25.37% of eight-dimensional space and the Leech lattice fills π¹²/12! ≈ 0.193% of twenty-four-dimensional space, and both numbers have been optimal, not merely record-holding, ever since. Viazovska received the Fields Medal in 2022.
Maryna Viazovska's The sphere packing problem in dimension 8 is fifteen pages long and constructs the certifying function outright. The two sign conditions pull in opposite directions, so she built the two halves separately: a +1 eigenfunction of the Fourier transform and a −1 eigenfunction, each written as an integral transform of a weakly holomorphic modular form, then combined so that the roots land exactly on √2, √4, √6, √8, … and nowhere else.
The modular forms are what supply the infinitely many roots for free. A quasi-periodicity in the upper half-plane becomes, after the transform, a function whose zeros are locked to the norms of an integral quadratic form — which is precisely the list of distances in E₈. Thirteen years of numerical searching had produced functions that came within a fraction of a percent; none of them could be pushed to equality by hand.
Drag the slider to demand more shells and watch the curve acquire another tangency it has to hit exactly. The real certificate has to do this at every shell, forever, on both the function and its Fourier transform at once. The vertical scale is compressed by default and the decaying envelope has been loosened deliberately — switch to linear and the later lobes all but disappear, which is what they genuinely do. These curves have the right roots and nothing else; Viazovska's are not elementary functions at all.
Try it: Drag Shells matched upward and count the constraints piling on. Three shells of E₈ already pin 9,120 vectors to exact equality; three shells of Leech pin over 415 million. Switch to the dimension-3 tab for the sobering version: the same picture can be drawn for face-centred cubic, and it leads nowhere — Cohn and Triantafillou proved in 2022 that the method is not sharp in dimensions 3, 4 or 5.
Nobody knows why 8 and 24 are the lucky dimensions, beyond the observation that they are where E₈ and Leech live — lattices so rigid and so symmetric that an inequality built for all packings happens to be exactly saturated by them. In every other dimension above three, the band you saw in the first demo is still open, and closing it will probably take an idea that does not exist yet.